For $\alpha$ a limit ordinal, show $V_{\alpha}=\bigcup_{\beta\lt\alpha}P(V_{\beta})$ For $\alpha$ a limit ordinal, I would like to show 


$V_{\alpha}=\bigcup_{\beta\lt\alpha}P(V_{\beta})$


where the $V$'s are members of the cumulative hierarchy and $P$ is the power set.
(This is a continuation of Showing equivalence of definitions for Zermelo Hierarchy, where Max ostensibly provided a solution in for a successor ordinal)
In the cumulative hierarchy, for a limit ordinal $\alpha$, $V_{\alpha}:=\bigcup_{\beta\lt\alpha}V_{\beta}$, so I would like to show 


$\bigcup_{\beta\lt\alpha}V_{\beta}=\bigcup_{\beta\lt\alpha}P(V_{\beta})$


Can I say that for any $\beta\lt\alpha$, there is a $\beta$' with $\beta\lt\beta'\lt\alpha$ such that $V_{\beta'}=P(V_{\beta})$. 
And conversely, for any such $\beta'$, there is a $\beta$ again with  $V_{\beta'}=P(V_{\beta})$, establishing inclusion in both dorections?
Thanks.
 A: In the following $\alpha$ is a limit ordinal (and $\alpha > 0$):
Your approach almost works. First of all, for all $\beta < \alpha$
$$
V_{\beta+1} = \mathcal P (V_\beta),
$$
so that $$\{ \mathcal P(V_\beta) \mid \beta < \alpha\} \subseteq \{ V_{\beta} \mid \beta < \alpha\} $$
The inverse inclusion does not hold, since for limit ordinals $\beta < \alpha$ (including $\beta = 0$) there is no $\gamma$ such that $V_{\beta} = \mathcal P(V_{\gamma})$.
But if $\beta < \alpha$ is a limit ordinal, then $V_{\beta} = \bigcup_{\gamma < \beta} V_{\gamma}$ which inductively yields that for all $x \in V_{\beta}$ there is some $\gamma < \beta$ such that $x \in V_{\gamma+1} = \mathcal P(V_{\gamma})$, so that
$$
\bigcup_{\beta < \alpha} V_\beta = \bigcup_{\beta < \alpha} \mathcal P(V_{\beta})
$$ 
after all.
A: The main idea for the reverse inclusion, I think, is that $V_\beta \subset P(V_\beta)$ because $V_\beta$ is transitive. 
For the first inclusion, what you said was enough, also perhaps you should say more clearly that (with your notations) $\beta' = \beta +1 $, so that $\beta < \alpha \implies \beta' < \alpha$ (as $\alpha$ is limit)
A: $b<a\implies b+1<a\implies P(V_b)=V_{b+1}\subset \cup_{c<a}V_c=V_a.$
So $\cup_{b<a}V_{b+1}\subset V_a.$ 
By transfinite induction every $V_b$ is a transitive set. If $x$ is transitive then $x\subset P(x).$ So $V_b\subset V_{b+1}.$ So $$V_a=\cup_{b<a}V_b\subset\cup_{b<a}V_{b+1}\subset V_a.$$
More generally if $S\subset a=\cup a$ then $\cup S=a\iff \cup_{b\in S}V_b=V_a.$
